Germs of continuous functions Let $R$ be the ring of germs of continuous functions $\mathbb R \rightarrow \mathbb R$.
It is clear that this is a local ring with maximal ideal $\mathcal m$ consisting of those germs $f$ with $f(0)=0$.
What is not clear to me is why $(\mathcal m)^2 =\mathcal m$ should hold.
Perhaps someone can give me a small hint how to show this.
 A: For $f\in \mathfrak m$ we have $f=\sqrt [3]f\cdot (\sqrt [3]f)^2$ and $\sqrt [3]f, (\sqrt [3]f)^2\in \mathfrak m$, so that indeed $\mathfrak m\subset \mathfrak m^2$.
The other inclusion is trivial and we have proved that $$\mathfrak m^2   = \mathfrak m $$ Edit: warning !
The analogue of the above result is false if one replaces the ring of germs of continuous functions by the ring of germs $\mathcal C_0^k\; (1\leq k\leq \infty)$ of $k$ times continuously differentiable  functions at the origin of $\mathbb R$.
Indeed if $f\in \mathfrak m^2$, we can write $f=\sum g_kh_k\; (g_k, h_k\in \mathfrak m)$  and differentiating yields $f'(0)=\sum g_k'(0)h_k(0)+\sum g_k'(0)h_k(0)=0$, so that elements $f\in \mathfrak m^2$ satisfy $f'(0)=0$ .
In particular $x\notin \mathfrak m^2$, although $x\in \mathfrak m$.
Hence $$\mathfrak m^2  \subsetneq \mathfrak m$$
A: The inclusion $(m)^2 \subset m$ is obvious. To prove the other inclusion, take $f \in m$. We wish to prove that $f \in (m)^2$. Define $f_+(x) = \max(f(x),0)$ and $f_-(x) = \min(f(x),0)$. Note that we have $f_+, f_- \in m$, $f_+ \ge 0$, $f_- \le 0$ and
$$f = f_+ + f_-$$
Now define $g = \sqrt{f_+}$ and $h = \sqrt{-f_-}$. We also have $h,g \in m$, and
$$f = g^2 - h^2 \in (m)^2$$
